Page 236 - Materials Chemistry, Second Edition

Groundwater Remediation 219

that the groundwater can enter at any level from the top to the bottom of
the aquifer.
− h )
Q = Kb h( 2 1 (AmericanPractical Units)
rr/)
528 log( 2 1
2.73 − h )
= Kb h ( 2 1 (SIUnits) (6.1)
rr /)
log( 2 1
where
Q = pumping rate or well yield (gpm or m /day)
3
h , h = static head measured from the aquifer bottom (ft or m)
1
2
r , r = radial distance from the pumping well (ft or m)
2
1
b = thickness of the aquifer (ft or m)
K = hydraulic conductivity of the aquifer (gpd/ft or m/day)
2
Example 6.1: Steady-State Drawdown from Pumping a Confined Aquifer
A confined aquifer 30-ft (9.1-m) thick has a piezometric surface 80 ft (24.4 m)
above the bottom confining layer. Groundwater is being extracted from a
4-in. (0.1-m) diameter fully penetrating well.
The pumping rate is 40 gpm (0.15 m /min). The aquifer is relatively sandy
3
and has a hydraulic conductivity of 200 gpd/ft . Steady-state drawdown of
2
5 ft (1.5 m) is observed in a monitoring well 10 ft (3.0 m) from the pumping
well. Determine

(a) The drawdown in the pumping well
(b) The radius of influence of the pumping well

Solution:
(a) First let us determine h (at r = 10 ft):
1
1
h = 80 − 5 = 75 ft (or = 24.4 − 1.5 = 22.9 m)
1
To determine the drawdown in the pumping well, set r at the
well = well radius = (2/12) ft = 0.051 m and use Equation (6.1):
(200)(30)( 2 h − 75)
40 = → h 2 = 68.74 ft
528 log[(2/12)/10]
or
2.73 [(200)(0.0410)](9.1)( 2 h − 22.9)
[(0.15)(1, 440)] = → h 2 = 21.0 m
log(0.051/3.0)
So the drawdown in the pumping well = 80 − 68.7 = 11.3 ft (or
= 24.4 − 21.0 = 3.4 m)